How do you find the asymptotes for $f(x)=(3x^2+2) / (x^2 -1)$ ?

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Answer 1 · 2016-01-21T10:47:05

The vertical asymptotes are $x=1$ and $x=-1$

The horizontal asymptote is $y=3$

The asymptotes occur where the denominator approaches zero, and where $x$ becomes very large, either positively or negatively.

$f(x) = (3x^2+2)/(x^2-1)$

In this case the denominator is the difference of two squares so the function can be rewritten as
$f(x)= (3x^2+2)/((x-1)(x+1))$

The denominator is zero when either $x=1$ or $x=-1$ and these are therefore the vertical asymptotes.

AS $x$ becomes very large, either positively or negatively,
$lim_(x->oo) =(3cancel(x^2) )/cancel(x^2)$

The horizontal asymptote is therefore $y=3$

How do you find the asymptotes for f(x)=(3x^2+2) / (x^2 -1)f(x)=(3x^2+2) / (x^2 -1)?